Q92 / 150 · MHT CET 2026 (15 April Shift 1) · Mathematics
The value of the integral $\int_{-\pi/2}^{\pi/2}\left(\dfrac{x^2 \cos x}{1+e^x}\right)dx$ is equal to $\left(\dfrac{\pi^2}{A}\right) - B$. Then $\left(\dfrac{A}{B}\right) = $
- A$-2$
- Correct answerB$2$
- C$6$
- D$-6$
Solution
Let $I = \int_{-\pi/2}^{\pi/2} \left(\dfrac{x^2 \cos x}{1+e^x}\right) dx$
Using the property $\int_{-a}^{a} f(x) dx = \int_{0}^{a} (f(x) + f(-x)) dx$, we get:
$I = \int_{0}^{\pi/2} \left( \dfrac{x^2 \cos x}{1+e^x} + \dfrac{(-x)^2 \cos(-x)}{1+e^{-x}} \right) dx$
$I = \int_{0}^{\pi/2} x^2 \cos x \left( \dfrac{1}{1+e^x} + \dfrac{e^x}{e^x+1} \right) dx$
$I = \int_{0}^{\pi/2} x^2 \cos x dx$
Applying integration by parts:
$I = [x^2 \sin x]_{0}^{\pi/2} - \int_{0}^{\pi/2} 2x \sin x dx$
$I = \dfrac{\pi^2}{4} - \left( [-2x \cos x]_{0}^{\pi/2} - \int_{0}^{\pi/2} (-2 \cos x) dx \right)$
$I = \dfrac{\pi^2}{4} - \left( 0 + [2 \sin x]_{0}^{\pi/2} \right)$
$I = \dfrac{\pi^2}{4} - 2$
Comparing this with $\dfrac{\pi^2}{A} - B$, we get $A = 4$ and $B = 2$.
Therefore, $\dfrac{A}{B} = \dfrac{4}{2} = 2$.
Answer: $2$
Using the property $\int_{-a}^{a} f(x) dx = \int_{0}^{a} (f(x) + f(-x)) dx$, we get:
$I = \int_{0}^{\pi/2} \left( \dfrac{x^2 \cos x}{1+e^x} + \dfrac{(-x)^2 \cos(-x)}{1+e^{-x}} \right) dx$
$I = \int_{0}^{\pi/2} x^2 \cos x \left( \dfrac{1}{1+e^x} + \dfrac{e^x}{e^x+1} \right) dx$
$I = \int_{0}^{\pi/2} x^2 \cos x dx$
Applying integration by parts:
$I = [x^2 \sin x]_{0}^{\pi/2} - \int_{0}^{\pi/2} 2x \sin x dx$
$I = \dfrac{\pi^2}{4} - \left( [-2x \cos x]_{0}^{\pi/2} - \int_{0}^{\pi/2} (-2 \cos x) dx \right)$
$I = \dfrac{\pi^2}{4} - \left( 0 + [2 \sin x]_{0}^{\pi/2} \right)$
$I = \dfrac{\pi^2}{4} - 2$
Comparing this with $\dfrac{\pi^2}{A} - B$, we get $A = 4$ and $B = 2$.
Therefore, $\dfrac{A}{B} = \dfrac{4}{2} = 2$.
Answer: $2$
Hence, the correct answer is option: 'B'.